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Can someone please help me with this integral $$\iint_D x^2y+y\sin(x^9) \ dxdy $$ $$D=\{(x,y):x^2+y^2\leq2,y>0\}$$

I have done like this: $$\int_{x=-\sqrt2}^{\sqrt2}\int_{y=0}^{\sqrt{2-x^2}} x^2y+y\sin(x^9) \ dydx$$ but the answer is wrong why?

Any suggestion would be great Thanks

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  • $\begingroup$ What answer did you get? Can you show us how? What answer have you been given? $\endgroup$
    – J.G.
    Jan 9, 2021 at 12:21
  • $\begingroup$ Just so you know the integral that you have set up also gives the correct answer. You may have made a mistake. It is not necessary that you have to integrate over $x$ first. $\endgroup$
    – Math Lover
    Jan 9, 2021 at 13:45

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$x$ should be integrated over first: $$\int_0^{\sqrt2}y\int_{-\sqrt{2-y^2}}^{\sqrt{2-y^2}}(x^2+\sin x^9)\,dx\,dy$$ Since $\sin x^9$ is an odd function and we are integrating over a domain symmetric about its oddness point, its contribution to the integral is zero and we may ignore it: $$=\int_0^{\sqrt2}y\int_{-\sqrt{2-y^2}}^{\sqrt{2-y^2}}x^2\,dx\,dy$$ $$=\int_0^{\sqrt2}y[x^3/3]_{-\sqrt{2-y^2}}^{\sqrt{2-y^2}}\,dy$$ $$=\frac23\int_0^{\sqrt2}y(2-y^2)^{3/2}\,dy$$ $$=-\frac13\int_0^{\sqrt2}-2y(2-y^2)^{3/2}\,dy$$ $$=-\frac13\int_2^0u^{3/2}\,du$$ $$=\frac13\cdot\frac25[u^{5/2}]_0^2$$ $$=\frac{2^{7/2}}{15}$$

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The $y\sin x^9$ part, being odd in $x$, doesn't contribute. Since $D$ has polar specification $0\le r\le\sqrt{2},\,0<\theta<\pi$, the integral is $\int_0^\pi\cos^2\theta\sin\theta d\theta\int_0^\sqrt{2}r^4dr$, which I leave you to evaluate.

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  • $\begingroup$ Thank you, does it matter if it was $y\geq0$ instead? $\endgroup$
    – simon
    Jan 9, 2021 at 12:33

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